Answer:
Plzzzzzzzzzzzzzzzz brainliest
Explanation:
In static friction, the frictional force resists force that is applied to an object, and the object remains at rest until the force of static friction is overcome. In kinetic friction, the frictional force resists the motion of an object. ... The frictional force itself is directed oppositely to the motion of the object.
Answer:
Speaking to witnesses is a job they do. If anyone stops after seeing the accident occur, the officers who respond to the accident take their personal information, they get their contact information, and they add their statements to their report.
Explanation:
Answer:
B. The buoyant force on the copper block is greater than the buoyant force on the lead block.
Explanation:
Given;
mass of lead block, m₁ = 200 g = 0.2 kg
mass of copper block, m₂ = 200 g = 0.2 kg
density of water, ρ = 1 g/cm³
density of lead block, ρ₁ = 11.34 g/cm³
density of copper block, ρ₂ = 8.96 g/cm³
The buoyant force on each block is calculated as;

The buoyant force of lead block;

The buoyant force of copper block

Therefore, the buoyant force on the copper block is greater than the buoyant force on the lead block
Answer:
The deceleration is
Explanation:
From the question we are told that
The distance of the car from the crossing is 
The speed is 
The reaction time of the engineer is 
Generally the distance covered during the reaction time is

=> 
=> 
Generally distance of the car from the crossing after the engineer reacts is
=>
=> 
Generally from kinematic equation

Here v is the final velocity of the car which is 0 m/s
So

=>
F=ma
for a Velocity/time
22/20
1.1
F=1.1. * 1100
F=1210newton